In $\Delta ABC$, $D$ is the midpoint of $AB$. $P$ is any point on $BC$, and $Q$ is a point on $AB$ such that $CQ \parallel PD$.

$PQ$ is joined in the given figure. Prove that $\text{Area}(\Delta BPQ) = \frac{1}{2} \text{Area}(\Delta ABC)$.                       [page No. 143]

Given:

D is the midpoint of ABP is any point on BC
Q is a point on AB
CQ || PD

$$\text{To prove: } \text{Area}(\Delta\text{BPQ}) = \frac{1}{2} \text{Area}(\Delta\text{ABC})$$

Step 1: Since D is midpoint of AB

$$AD = DB$$

So, $AB = 2DB$

Step 2: Use the property of parallel lines

Since CQ || PD.

Triangles $\Delta$BPD and $\Delta$BCQ are similar.

$$\text{Therefore, } \frac{BD}{BQ} = \frac{BP}{BC}$$

$$\text{But, }BD=\frac{AB}{2}$$

$$\text{Hence, } \frac{AB}{2BQ} = \frac{BP}{BC}$$

Cross-multiplying: AB . BC = 2(BQ . BP)

Step 3: Compare areas

$$\text{Area of } \Delta\text{ABC} = \frac{1}{2} (\text{AB})(\text{height from C})$$

$$\text{Area of } \Delta\text{BPQ} = \frac{1}{2} (\text{BQ})(\text{height from P})$$

Since both heights correspond proportionally from the similar triangles relation obtained above, we get

$$2 \text{Area}(\Delta\text{BPQ}) = \text{Area}(\Delta\text{ABC})$$

Hence, proved.